# Top person sorted by score

The Prover-Account Top 20
Persons by: number score normalized score
Programs by: number score normalized score
Projects by: number score normalized score

At this site we keep several lists of primes, most notably the list of the 5,000 largest known primes. Who found the most of these record primes? We keep separate counts for persons, projects and programs. To see these lists click on 'number' to the right.

Clearly one 100,000,000 digit prime is much harder to discover than quite a few 100,000 digit primes. Based on the usual estimates we score the top persons, provers and projects by adding ‎(log n)3 log log n‎ for each of their primes n. Click on 'score' to see these lists.

Finally, to make sense of the score values, we normalize them by dividing by the current score of the 5000th prime. See these by clicking on 'normalized score' in the table on the right.

rankpersonprimesscore
21 Ben Maloney 1 52.0371
22 Frank Matillek 10 52.0287
23 Wolfgang Schwieger 86 51.9175
25 Marc Wiseler 9 51.8176
26 Diego Bertolotti 1 51.6397
27 Rudi Tapper 4 51.6208
28 Randall Scalise 156 51.3752
29 Brian D. Niegocki 21 51.3426
30 Hiroyuki Okazaki 46 51.1617
31 Peter Kaiser 84.3333 51.0046
32 Alen Kecic 17 50.9575
33 Michael Cameron 1 50.9234
34 Erik Veit 48 50.9066
35 Thomas Ritschel 77 50.8523
36 Konstantin Agafonov 1 50.8197
37 Vaughan Davies 53 50.7222
38 Ken Glennie 2 50.6508
39 Peter Benson 75 50.6262
40 Honza Cholt 24 50.5525

#### Notes:

Score for Primes

To find the score for a person, program or project's primes, we give each prime n the score (log n)3 log log n; and then find the sum of the scores of their primes. For persons (and for projects), if three go together to find the prime, each gets one-third of the score. Finally we take the log of the resulting sum to narrow the range of the resulting scores. (Throughout this page log is the natural logarithm.)

How did we settle on (log n)3 log log n? For most of the primes on the list the primality testing algorithms take roughly O(log(n)) steps where the steps each take a set number of multiplications. FFT multiplications take about

O( log n . log log n . log log log n )

operations. However, for practical purposes the O(log log log n) is a constant for this range number (it is the precision of numbers used during the FFT, 64 bits suffices for numbers under about 2,000,000 digits).

Next, by the prime number theorem, the number of integers we must test before finding a prime the size of n is O(log n) (only the constant is effected by prescreening using trial division).  So to get a rough estimate of the amount of time to find a prime the size of n, we just multiply these together and we get

O( (log n)3 log log n ).

Finally, for convenience when we add these scores, we take the log of the result.  This is because log n is roughly 2.3 times the number of digits in the prime n, so (log n)3 is quite large for many of the primes on the list. (The number of decimal digits in n is floor((log n)/(log 10)+1)).